Compute line integral $\int_a^b (y^2z^3dx + 2xyz^3dy + 3xy^2z^2dz)$

Question

Compute line integral $\int_a^b (y^2z^3dx + 2xyz^3dy + 3xy^2z^2dz)$ where $a = (1,1,1)$ and $b = (2,2,2)$

What I have done:

To find $t$ I used the calculation for slope:

$\frac{x-1}{2-1}=t, \frac{y-1}{2-1}=t, \frac{z-1}{2-1}=t$ and then re-arranged for x,y,z to calculate the derivate to find that $dx =1, dy=1, dz=1$ and $t+1 = (x, y, z)$

Plugging this back into the integral: $$6\int_a^b (t+1)^5 dt$$

However, what do I do with the integral bounds? Do I set them from $t \in [1,2]$?

Answer 1

The hint you mentioned implies that you can use the fundamental theorem of calculus. Note that $$ (y^2z^3,2xyz^3,3xy^2z^2)=\nabla f(x,y,z) $$ where $f(x,y,z)=xy^2z^3$. So the integral is path independent, and you don't really need to parametrize your path: $$ \int_A^B \nabla f(x,y,z)\cdot dr=f(B)-f(A) $$ where $A=(1,1,1)$, $B=(2,2,2)$.

Usually, the notation $\int_A^B$ is only used for path integrals that are path independent. Otherwise, one usually uses $\int_\gamma$ instead and describes $\gamma$ in words.

Answer 2

Since $(x,y,z)=(t+1,t+1,t+1)$ and you want to go from $(1,1,1)$ to $(2,2,2)$, you should compute that integral with $a=0$ and $b=1$.

Note that talking about $\int_a^b(y^2z^3\,\mathrm dx+2xyz^3\,\mathrm dy+3xy^2z^2\,\mathrm dz)$ only makes sense because your vector field is conservative. Otherwise, the integral would depend upon the path going from $(1,1,1)$ to $(2,2,2)$.